\(m_{H_2O}=1752\cdot1=1752\left(g\right)\)

\(n_{Na_2O}=\dfrac{248}{62}=4\left(mol\right)\)

\(Na_2O+H_2O\rightarrow2NaOH\)

\(4...............................8\)

\(m_{NaOH}=8\cdot40=320\left(g\right)\)

\(m_{dd}=248+1752=2000\left(g\right)\)

\(C\%_{NaOH}=\dfrac{320}{2000}\cdot100\%=16\%\)

PTHH: \(Na_2O+H_2O\rightarrow2NaOH\)

Ta có: \(n_{NaOH}=2n_{Na_2O}=2\cdot\dfrac{248}{62}=8\left(mol\right)\)

\(\Rightarrow C\%_{NaOH}=\dfrac{8\cdot40}{248+1752}\cdot100\%=16\%\)

\(n_{Na_2O}=\dfrac{15,5}{62}=0,25\left(mol\right)\\ PTHH:Na_2O+H_2O\rightarrow2NaOH\\ n_{NaOH}=2.0,25=0,5\left(mol\right)\\ a,C_{MddNaOH}=\dfrac{0,5}{0,5}=1\left(M\right)\\ b,2NaOH+H_2SO_4\rightarrow Na_2SO_4+2H_2O\\ n_{H_2SO_4}=n_{Na_2SO_4}=\dfrac{0,5}{2}=0,25\left(mol\right)\\ m_{H_2SO_4}=0,25.98=24,5\left(g\right)\\ m_{ddH_2SO_4}=\dfrac{24,5.100}{20}=122,5\left(g\right)\\ V_{ddH_2SO_4}=\dfrac{122,5}{1,14}\approx107,456\left(ml\right)\\ c,V_{ddsau}=V_{ddNaOH}+V_{ddH_2SO_4}\approx0,5+0,107456=0,607456\left(l\right)\\C_{MddNa_2SO_4}\approx\dfrac{ 0,25}{0,607456}\approx0,411552\left(M\right)\)

Bài 2

\(C_{\%đường}=\dfrac{10}{10+100}\cdot100\%\approx9,09\%\)

Bài 3

\(n_{NaOH}=\dfrac{4}{40}=0,1mol\\ C_{M_{NaOH}}=\dfrac{0,1}{0,2}=0,5M\)

\(m_{NaOH\left(A\right)}=20.5\%=1\left(g\right)\)

Trong B:  

gọi x là khối lượng Na2O thêm vào , x>0 (g)

\(10\%=\dfrac{\dfrac{80}{62}x+1}{x+20}\)

\(\rightarrow x=0,84\left(g\right)\)

Vậy khối Na2O thêm vào dd A là 0,84 (g) 

b, \(m_{KOH\left(A\right)}=2\%.20=0,4\left(g\right)\)

\(C\%_{KOH\left(B\right)}=\dfrac{0,4}{20+0,84}.100\%=1,92\%\)

nNa2O= 12,4/62=0,2(mol)

a) PTHH: Na2O + H2O -> 2 NaOH

0,2_________0,2____0,4(mol)

b) VddNaOH=2(l)

=>CMddNaOH=0,4/0,2=2(M)

Chúc em học tốt!

giup mk voi

$Al_2O_3+3H_2SO_4\rightarrow Al_2(SO_4)_3+3H_2O$

$n_{Al_2O_3}=\dfrac{10,2}{102}=0,1\ mol$

Theo phương trình:

$n_{H_2SO_4}=3n_{Al_2O_3}=0,3\ mol$

$n_{Al_2(SO_4)_3}=n_{Al_2O_3}=0,1\ mol$

a) Khối lượng muối thu được:

$m_{Al_2(SO_4)_3}=0,1\cdot342=34,2\ gam$

b) Nồng độ phần trăm dung dịch $H_2SO_4$:

$m_{H_2SO_4}=0,3\cdot98=29,4\ gam$

$C\%_{H_2SO_4}=\dfrac{29,4}{200}\cdot100\%=14,7\%$

c) Nồng độ phần trăm dung dịch muối:

$m_{dd\ sau\ phản\ ứng}=200+10,2=210,2\ gam$

$C\%_{Al_2(SO_4)_3}=\dfrac{34,2}{210,2}\cdot100\%\approx16,27\%$

Vậy:

$a)\ m_{muối}=34,2\ gam$

$b)\ C\%_{H_2SO_4}=14,7\%$

$c)\ C\%_{Al_2(SO_4)_3}\approx16,27\%$

=50/400.100%=12,5%.

Em đăng vào môn Hoá hấy

C%=50/400.100%=12,5%.